Question

In Fig. E is a point on side CB produced of an

isosceles triangle ABC with AB = AC. If AD ⊥

BC and EF ⊥AC, prove that ∆ ABD ~ ∆ ECF.​

Answer 1

It is given that △ABC is isosceles with

AB=AC

∴ ∠B=∠C

Now, in ∆′sABD and ECF, we have

∠ ABD=∠ECF [∵∠B=∠C]

∠ADB=∠EFC=90⁰ [∵AD⊥BC and EF⊥AC]

So, by AA-criterion of similarity, we have

△ABD∼△ECF [Hence proved✅]

Answer 2

Answer: It is given that △ABC is isosceles with

AB=AC

∴ ∠B=∠C

Now, in ∆′sABD and ECF, we have

∠ ABD=∠ECF [∵∠B=∠C]

∠ADB=∠EFC=90⁰ [∵AD⊥BC and EF⊥AC]

So, by AA-criterion of similarity, we have

△ABD∼△ECF [Hence proved✅]

Step-by-step explanation: